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telo118 [61]
3 years ago
9

An electric field can be created by a single charge or a distribution of charges. The electric field a distance from a point cha

rge has magnitude E = k|q'|/r^2.
The electric field points away from positive charges and toward negative charges. A distribution of charges creates an electric field that can be found by taking the vector sum of the fields created by individual point harges. Note that if a charge is placed in an electric field created by q', q will not significantly affect the electric field if it is small compared to q'. Imagine an isolated positive point charge with a charge Q (many times larger than the charge on a single electron).
1. There is a single electron at a distance from the point charge. On which of the following quantities does the force on the electron depend?
a. the distance between the positive charge and the electron
b. the charge on the electron
c. the mass of the electron
d. the charge of the positive charge
e. the mass of the positive charge
f. the radius of the positive charge
g. the radius of the electron
2. For the same situation as in Part A, on which of the following quantities does the electric field at the electron's position depend?
a. the distance between the positive charge and the electron
b. the charge on the electron
c. the mass of the electron
d. the charge of the positive charge
e. the mass of the positive charge
f. the radius of the positive charge
g. the radius of the electron
Physics
1 answer:
Reika [66]3 years ago
7 0

Answer:

a) true.

b) True

c) False. In the equation above the mass does not appear

d) True

e) False. Mass does not appear in the equation

f) False. The load even when distributed in the space can be considered concentrated in the center

Explanation:

1. The electric force is given by the relation

           F = k Q e / r2

where k is the Coulomb constant, Q the charge used, e the charge of the electron and r the distance between the two.

 The strength depends on:

a) true.

b) True

c) False. In the equation above the mass does not appear

d) True

e) False. Mass does not appear in the equation

f) False. The load even when distributed in the space can be considered concentrated in the center

two.

a) True

b) Treu

c) Fail

f) false

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The bohr model of the atom addressed the problem of
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Rutherford's nuclear model stated that a cloud of negative electrons surround the nucleus however scientists realized that electrons in a cloud around the nucleus of an atom would be attracted to the nucleus, causing the atom to collapse. Thus Bohr's model proposed that electrons were contained in shells and they orbit the nucleus at fixed distances.
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3 years ago
A car of mass m=1000kg is traveling at speed v and brakes. The skid marks are 20m long and the coefficient of kinetic friction i
scoray [572]

Answer:

v = 14 m/s

  = 31.3 mph

The answer would be the same if the mass of the car were 2000 kg

Explanation:

Let V be the final velocity of the car after skidding, and v be the initial velocity of the car. Let a be the acceleration of the car and Δx be the distance the car travels after applying brakes (length of the skid marks). Let Fk be the force of friction between the tyres and the road. Let N be the normal force exerted on the car and μ be the co efficient of kinetic friction.

V^2 = v^2 + 2×a×Δx

Now V, the final velocity is zero as the car stops

0 = v^2 + 2×a×Δx

v^2 = -2×a×Δx

v =√-2×a×Δx    .....*

Now applying Newton's Second Law

Fnet = m×a

-Fk = m×a

-μ×N = m×a

-μ×m×g = m×a (The mass cancels out)

a =  -μ×g

Substituting the value of a back to equation *

v = √-2×(-μ×g)×Δx  

v = √-2×(-0.5×9.8)×20

v = 14 m/s

Therefore the speed the car was travelling with v = 14 m/s

which is equal to 31.3 mph

Now if you were to change the mass of the car to 2000 kg the value for v would still be the same. As it is seen above mass cancels out so it does not influence or affect the value of the velocity obtained.

8 0
3 years ago
A bucket of water with total mass 23 kg is attached to a rope, which in turn, is wound around a 0.050-m radius cylinder at the t
aliina [53]

The question seems a bit incomplete. The question should be as follow:

A bucket of water with total mass 23 kg is attached to a rope, which in turn, is wound around a 0.050-m radius cylinder at the top of a well. A crank with a turning radius of 0.25 m is attached to the end of the cylinder. What minimum force directed perpendicular to the crank handle is required to just raise the bucket? (Assume the rope's mass is negligible, that cylinder turns on frictionless bearings, and that g= 9.8 m/s2.)

Answer:

45.08N

Explanation:

The question involves moment topic. Note that the equation of moment, M is

Moment, M = Force, F x Perpendicular distance from the turning point, d

M = F x d

Moment involves turning movement along the pivot. in this case, we have 2 pivots. The first one the attached near the rope, while the other is the crank.

To raise the bucket, minimum force required must be able to provide at least the same moment as the moment due to the bucket of water. i.e.

Moment due to bucket = moment due to force on the crank

First find the moment due to bucket,

Mb = F (weight) x d (radius of cylinder on top of well)

   =  (23 x 9.8) x 0.05

   = 11.27 Nm

Next, Find the moment due to force on the crank. Note that the force is the minimum required force for this equation.

Mc = F (min Force) x d (radius of crank)

     = F x  0.25  = 0.25F

Now we get a simple equation of

11.27 = 0.25F

Using Algebra, we'll get the minimum Force, F:

F = 11.27 / 0.25

  = 45.08 N

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4 years ago
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Explanation:

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